# Intervals

Closed Interval: Let a and b be two given real numbers such that a < b. Then the set of all real numbers x such that a ≤ x ≤ b is called a closed interval and is denoted by [a, b] i.e., [a, b] = {x ϵ R| a ≤ x ≤ b}

Open Interval: Let a and b be two given real numbers such that a < b. Then the set of all real numbers x such that a ≤ x ≤ b is called a closed interval and is denoted by (a, b) i.e., (a, b) = {x ϵ R| a ≤ x ≤ b}

Semi-Closed OR Semi Open Interval: If a, b are two given real numbers such that a < b then the sets (a, b) = {x ϵ R| a ≤ x ≤ b} and (a, b) = {x ϵ R| a ≤ x < b} are known as semi-closed or semi-open intervals and are also denoted by ] a, b [ and [a, b] respectively.

REAL FUNCTIONS:

Real Function: If the domain and co-domain of a function are subsets of R (set of all real numbers). It is called a real valued function or in short a real function.

Description of Areal Function: If f is a real valued function with finite domain, then f can be described by listing the values which it attains at different points of its domain. However, if the domain of a real function is an infinite set, then, f cannot be described by listing the values at points in its domain. In such cases real functions are generally described by some general formula or rule like f (x) = x2 + 1 or f (x) = 2sinx + 3 etc.

Example: If $$f\left( x \right)=x+\frac{1}{x}$$, prove that $${{\left[ f\left( x \right) \right]}^{3}}=f\left( {{x}^{3}} \right)+3f\left( \frac{1}{x} \right)$$.

Solution: We have,

$$f\left( x \right)=x+\frac{1}{x}$$.

⇒ $$f\left( {{x}^{3}} \right)={{x}^{3}}+\frac{1}{{{x}^{3}}}$$.

Now,

$${{\left[ f\left( x \right) \right]}^{3}}={{\left( x+\frac{1}{x} \right)}^{3}}$$.

$$=\left( {{x}^{3}}+\frac{1}{{{x}^{3}}} \right)+3\left( x+\frac{1}{x} \right)$$.

= f (x3) + 3 f (x) .

$$=f\left( {{x}^{3}} \right)+3f\left( \frac{1}{x} \right)\,$$      [∵ f (x) = f(1/x)].

DOMAIN AND RANGE OF A REAL FUNCTION:

Domain: Generally real functions in calculus are described by some formula and their domains are not explicitly stated. In such cases to find the domain of a function f (say) we use the fact that the domain is the set of all real numbers x for which f (x) is a real number.

Range: As discussed, the range of a function f (x) is the set of values of f (x) which it attains at points in its domain. For a real function the co-domain is always a subset of R. So, range of a real function f is the set of all points  such that y = f (x) where x ϵ Dom f (x)

ALGORITHM:

Step 1: Put f (x) = y

Step 2: Solve the equations in step 1 for x to obtain x = ϕ (y).

Step 3: Find the values of y for which the values of x, obtained from x = ϕ (y) are in the domain of f.

Step 4: The set of values of y obtained in step 3 is the range of f.

Example: Find the domain and range of the function $$f\left(x\right)=\frac{1}{2-\cos 3x}$$.

Solution: We have, $$f\left( x \right)=\frac{1}{2-\cos 3x}$$.

Domain: We have,

-1 ≤ cos 3x ≤ 1 for all x ϵ R,

⇒ -1 ≤ -cos3x ≤ 1

⇒ -1 ≤ 2 – cos 3x ≤ 3 for all x ϵ R,

⇒ f (x) is defined for all x ϵ R

So, domain (f) = R.

Range: Let f (x) = y Then

⇒ $$\frac{1}{2-\cos 3x}=y$$.

⇒ $$2-\cos 3x=\frac{1}{y}$$.

⇒ $$\cos 3x=2-\frac{1}{y}$$.

⇒ $$-1\le 2-\frac{1}{y}\le 1\,$$   [∴ – 1 ≤ cos3x ≤ 1].

⇒ $$-3\le -\frac{1}{y}\le -1\,$$.

⇒ 3 ≥ 1/y ≥ 1

⇒ ⅓ ≤ y ≤ 1

⇒ y ϵ [⅓, 1]

So, range of f = [⅓, 1].

Some Standard Real Functions:

Constant Function: Let k be a fixed real number. Then a function f (x) given by f (x) = k for all x ϵ R is called a constant function. Sometimes we also call it the constant function k.

Identity Function: The function defined by I (x) = x for all x ϵ R, is called the identity function on R.

Modulus Function: The function defined by f (x) = |x| =\left\{\begin{align}& x,\,\,\,when\,\,\,x\ge 0 \\& -x,\,\,when\,\,\,x<0 \\\end{align} \right. is called the modulus function.

Properties of Modulus Function: The modulus function has the following properties:

i) For any real number x we have √x2 = |x|.

ii) If a, b are positive real numbers then

⇒ x2 ≤ a2 |x| ≤ a ⇔ -a ≤ x ≤ a

⇒ x2 ≥ a2 ⇔|x| ≥ a ⇔ x ≤ -a or x ≥ a

⇒ x2 > a2 ⇔|x| > a ⇔ x < -a or x > a

⇒ a² ≤ x² ≤ b² ⇔ a ≤ |x| ≤ b ⇔ x ϵ [-b, -a] U [a, b]

⇒ a2 ≤ x2 ≤ b2 ⇔ a ≤ |x| ≤ b ⇔ x ϵ [-b, -a] U [a, b]

iii) |x + y| = |x| + |y|

⇒ (x ≥ 0 and y ≥ 0) or (x < 0 and y < 0)

iv) |x – y| = |x| – |y|

⇒ (x ≥ 0) and |x| ≥ |y| or (x ≤ 0, y ≤ 0 and |x| ≥ |y|)

v) |x ± y| < |x| + |y|

vi) |x ± y| ≥ ||x| – |y||